Exponential Functions/Logarithmic Functions

Definition of Exponents Let $a$ be a positive real number, $n$ be a natural number, $m$ be an integer. Then

1. $a^{0} = 1$
2. $a^{n} = \overbrace{a\cdot a \cdots a}^{n}$
3. $a^{-n} = \frac{1}{a^{n}}$
4. $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$
5. Let $\alpha$   be irrational number, $\alpha = \lim_{n \to \infty}\alpha_{n}$, where $\{\alpha_{m}\}$   is sequence of rational number. Then $a^{\alpha} = \lim_{n \to \infty}a^{\alpha_{n}}$

Definition of Exponents To define the exponents, $a$ has to be positive.
Note that $\frac{1}{a}\cdot a = \frac{a}{a} = 1$, and $a^{-1}\cdot a = a^{-1+1} = a^0 = 1$. Thus $a^{-1} = \frac{1}{a}$

NOTE At 5. Consider the case $a \geq 1$ and $a^{\sqrt{2}}$. First we create an increasing sequence of rational numbers which converges to $\sqrt{2}$, say $\{\alpha_{n}\}$. For example,

$$\{1.4,1.41,1.414,\ldots\}.$$

Then for $m < n$, we have $\alpha_{m} < \alpha_{n}$ and

$$\frac{a^{\alpha_{n}}}{a^{\alpha_{m}}} = a^{\alpha_{n}-\alpha_{m}} > a^{0} = 1.$$

Thus, $\{a^{\alpha_{n}}\}$ is increasing . Also $a^{\alpha_{n}} \leq a^{\sqrt{2}}$ implies that

$\{a^{\alpha_{n}}\}$ is a bounded above increasing sequence. Thus it converges. Let

$$a^{\sqrt{2}} = \lim_{n \rightarrow \infty} a^{\alpha_{n}}$$

Similarly for $0 < a < 1$. The proof looks OK. But there are many rational valued sequences which converge to $\sqrt{2}$. So, we have to show using other rational sequence $b_n$ which converges to $\sqrt{2}$, $\lim_{n \rightarrow \infty}a^{b_{n}}$ is the same. In mathematics, we call this Uniqueness.

Check

Since $a > 1$, $\alpha_{n} \leq \sqrt{2}$ implies $a^{\alpha_{n}} \leq a^{\sqrt{2}}$.

Let $\{\alpha_{n}\},\{\beta_{n}\}$ be two increasing rational valued sequences which converege to $\sqrt{2}$. Since $\alpha_{n} \rightarrow \sqrt{2}, \beta_{n} \rightarrow \sqrt{2}$, for all $\beta_{n}$ satisfying $\alpha_{1} \leq \beta_{n}$, we can choose $\alpha_{m},\alpha_{m+1}$ so that $\alpha_{m}\leq \beta_{n} < \alpha_{m+1}$. $\alpha_{n}, \beta_{n}$ are both increasing and converge $\sqrt{2}$. Thus we can choose $\beta_{n}$ which is larger that $\alpha_{1}$. Then we can choose $\alpha_{k}$ which is larger that $\beta_{n}$. Thus we cha choose $m$ so that $\alpha_{m} \leq \beta_{n} < \beta_{m+1}$.

For $a > 1$,

$$a^{\alpha_{m}} \leq a^{\beta_{n}} \leq a^{\alpha_{m+1}}$$

and for $0 < a < 1$,

$$a^{\alpha_{m}} \geq a^{\beta_{n}} > a^{\alpha_{m+1}}$$

Now $n \rightarrow \infty$ implies that $m \rightarrow \infty$ and by the squeezing theorem,

$$\lim_{m \rightarrow \infty}a^{\alpha_{m}} = \lim_{n \rightarrow \infty}a^{\beta_{n}}$$

This shows that $a^{\sqrt{2}}$ is independent from the choice of $\{a_{n}\}$.

For $a > 0$ and $a \neq 1$, a function $y = a^{x}$ is called Exponential Function .

The domain is $(-\infty,\infty)$ and the range is $(0, \infty)$.

Laws of Exponential Function

Theorem 1..13   Suppose that $a > 0$. Then for every real numbers $x,y$, we have

1. $a^0 = 1$
2. $a^{x+y} = a^{x}a^{y}$
3. $a^{x-y} = \frac{a^{x}}{a^{y}}$
4. $(a^{x})^{y} = a^{xy}$
5. $e^{x}$ is continuous on $(-\infty,\infty)$ and strictly increasing function.
6. $\lim_{x \rightarrow \infty}e^{x} = \infty, \lim_{x \rightarrow -\infty}e^{x} = 0$

NOTE $a^{x}$ for $a > 0$ is defined for irrational number $x$ by considering $a^{\alpha}$.

$$a^{\alpha} = \lim_{n \rightarrow \infty}a^{\alpha_{n}}, \alpha_{n} \rightarrow \alpha,\{\alpha_{n}\}$$sequence of rational numbers $$, a > 0$$

Then the case 2., we have $a^x = \lim_{n \rightarrow \infty}a^{\alpha_{n}}, a^y = \lim_{n \rightarrow \infty}a^{\beta_{n}}$ and

$$a^{x}a^{y} = \lim_{n \rightarrow \infty}a^{\alpha_{n}} \cdot \lim... ...{\beta_{n}} = \lim_{n \rightarrow \infty}a^{\alpha_{n} + \beta_{n}} = a^{x+y}.$$

Graph of Exp Among all exponential functions $a^x$, $e^x$ is the most important function.

Example 1..30   Draw the graph of $y = 2^{x}$ and $y = (\frac{1}{2})^{x}$.

Figure 1.47: Example1-30

NOTE We find values of $x$ and corresponding values of $y$. Then plot those points and connect by smooth curve.

$$\begin{array}{c\vert cccccccccccccccc}\hline x & \cdots & -3 ... ...frac{1}{4} & \cdots & \frac{1}{8} & \cdots\ \hline \end{array}$$

Exercise 1..30   Suppse that $f(x)$ satisfies $f(x+y) = f(x)f(y)$ for all $x,y$ and $f(0) \neq 0$. Show the followings.
1. $f(0) = 1$ 2. $f(-y) = \frac{1}{f(y)}$

The property $f(x+y) = f(x)f(y)$ is generalization of exponetial function.

SOLUTION 1. We can write $f(0) = f(0+0) = f(0)f(0)$. Then $f(0)(f(0) - 1) = 0$. Since $f(0) \neq 0$, we have $f(0) = 1$ $\blacksquare$
2. $1 = f(0) = f(y - y) = f(y)f(-y)$ implies that $f(-y) = \frac{1}{f(y)}$ $\blacksquare$

Definition of Logarithm Let $a$ be a positive real number and $a \neq 1$. The for every real number $x$ and $y = a^{x}$, we write $x = \log_{a}{y}$ and call $a$ base of Logarithm.

Table of Logarithm Table of logarithm of base 2. NOTE

$$\begin{array}{\vert c\vert c\vert c\vert c\vert c\vert c\vert... ...6 & 7\ \hline y & 2 & 4 & 8 & 16 & 32 & 64 & 128 \end{array}$$

Calculate $8 \times 16$. Find the number above $8$ and the number above $16$. Then we have 3 and 4. Now add these two numbers to obtain $7$. Then the numer below $7$ is $128$ the result of $8 \times 16$. Express the number 3 above 8 as $\log_{2}{8}$. Then we get $\log_{2}(8 \times 16) = \log_{2}{8} + \log_{2}16 = 3 + 4 = 7$. Next we calclulate $128 \div 32$. This time we subtract the number above 128 from the number above 32. Then we get $7 - 5 = 2$. Now the number 4 is the result of $128 \div 32$. Thus $\log_{2}(128 \div 32) = \log_{2}{128} - \log_{2}{32} = 7 - 5 = 2$.

Laws of Logarithmics

Theorem 1..14   Let $a > 0$, $a \neq 1$, $p > 0, q > 0$. Then we have the followings.

1. $\log_{a}{pq} = \log_{a}p + \log_{a}q$
2. $\log_{a}\frac{p}{q} = \log_{a}p - \log_{a}q$
3. $\log_{a}\frac{1}{q} = -\log_{a}q$
4. $\log_{a}q^r = r\log{a}q$

Laws of Logarithms Take logarithms Then
a product becomes an addition,
a quotient becomes a subtraction,
a power becomes a product.

NOTE Consider $y = \log_{a}{x}$. Then by the definition of logarithm, we have $x = a^{y}$. Now take the logarithm of both sides with the base $b$. Then

$$\log_{b}x = \log_{b}a^{y} = y\log_{b}a$$

From this, we have

$$y = \frac{\log_{b}x}{\log_{b}a}.$$

We call this change-of-base formula.

Existence of Inverse Function Strictly increasing function is one-to-one. Thus, we have a inverse function. Natural Logarithm We write natural logarithm without the base $b$.

The domain of $y = a^{x}$ is $(0, \infty)$, and $y = a^x$ is strictly increasing and continuous function. Thus there exists a unique inverse function and we write $x = f^{-1}(y) = \log_{a}{y}$. Note $\log_{a}{y}$ is defined and continuous on $(0, \infty)$ . we say $\log_{a}{y}$ logarithmic function with base $a$.

Laws of Logarithmic Functions

Theorem 1..15   Let $a > 0$. Then for every real numer $x > 0,y > 0$, we have the followings.

1. $\log_{a}{1} = 0, \log{e} = 1$
2. $\log_{a}(xy) = \log_{a}{x} + \log_{a}{y}$
3. $\log_{a}\frac{x}{y} = \log_{a}{x} - \log_{a}{y}$
4. $\log_{a}x^{y} = y\log_{a}{x}$
5. $\log{x}$ is continuous and monotonically increasing on $(0, \infty)$
6. $\lim_{x \rightarrow \infty}\log{x} = \infty,\lim_{x \rightarrow 0+}\log{x} = -\infty$

Figure 1.48: Graph of Logarithm

NOTE To show 2. Let $\alpha = \log_{a}x, \beta = \log_{a}y$. Then $x = a^{\alpha}, y = \alpha^{\beta}$ and

$$xy = a^{\alpha}a^{\beta} = a^{\alpha + \beta}$$

Thus,

$$\log_{a}x + \log_{a}y = \alpha + \beta = \log_{a}{xy}$$

Example 1..31   Find the domain of the following function.

$$f(x) = \log{(x\sqrt{1+3x})}$$

Figure 1.49: Example1-31

A logarithmic function can not take the values less than 0. SOLUTION Since a logarithmic function can only take positive values, we have $x\sqrt{1+3x} > 0$. Thus,

$$x > 0$$   and$$\sqrt{1+3x} > 0$$   or$$x < 0$$   and$$\sqrt{1+3x} < 0.$$

$\sqrt{1+3x} \geq 0$ implies $x > 0$   and$\sqrt{1+3x} > 0$. Since a square root function can not take negative values, we have $1 + 3x > 0$. Thus $x > 0$ and $x > -\frac{1}{3}$. Express using interval, we have $(0,\infty) \cap (-\frac{1}{3},\infty) = (0, \infty)$ $\blacksquare$

Exercise 1..31   Suppose that $f(x)$ satisfies $f(xy) = f(x)+ f(y)$ for $x > 0,y > 0$. Then
1. $f(1) = 0$ 2. $f(\frac{1}{y}) = -f(y)$

The property $f(xy) = f(x)+ f(y)$ is generalization of logarithm.

SOLUTION 1. $f(1) = f(1\cdot 1) = f(1) + f(1)$ implies $f(1) = 0$ $\blacksquare$
2. $0 = f(1) = f(\frac{y}{y}) = f(y \cdot \frac{1}{y}) = f(y) + \frac{1}{f(y)}$ implies that $f(\frac{1}{y}) = -f(y)$ $\blacksquare$

Hyperbolic Functions

A function below is called hyperbolic function.

$$\sinh{x} = \frac{e^{x} - e^{-x}}{2}, \cosh{x} = \frac{e^{x} + e... ... \tanh{x} = \frac{\sinh{x}}{\cosh{x}} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}$$

The curve described by a uniform chain hanging from two supports in a uniform gravitational field is called a catenary. To exptess catenary, $\cosh{x}$ is used.

Example 1..32   Show $${\lim_{x \rightarrow \infty}(1 + \frac{1}{x})^x = e}$$ .

SOLUTION As $x \rightarrow \infty$, there exists $n$ which satisfies $n < x < n+1$. Thus Calculation Take the reciprocal of $n < x < n+1$. Then we have $\frac{1}{n} > \frac{1}{x} > \frac{1}{n+1}$. Now add 1 to both sides of inequality. $\frac{1}{n}+ 1 > \frac{1}{x} + 1 > \frac{1}{n+1} + 1$
Also, $(1 + \frac{1}{n+1})^n < (1 + \frac{1}{n+1})^x < (1 + \frac{1}{x})^x$

$$1 + \frac{1}{n+1} < 1 + \frac{1}{x} < 1 + \frac{1}{n}$$

Use $n < x < n+1$ one more time to get

$$(1 + \frac{1}{n+1})^n < (1 + \frac{1}{x})^x < (1 + \frac{1}{n})^{n+1}$$

Note that

$$(1 + \frac{1}{n+1})^n = \frac{(1 + \frac{1}{n+1})^{n+1}}{(1 + \fr... ...}{n+1})}, (1 + \frac{1}{n})^{n+1} = (1 + \frac{1}{n})^{n} (1 + \frac{1}{n}).$$

Then

$$e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n+1})^n \leq \lim_{... ... + \frac{1}{x})^x \leq \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^{n+1} = e.$$

By the squeezing theorem

$$\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e\ensuremath{ \blacksquare}$$

We already know $\lim_{n \to \infty}(1 + \frac{1}{n})^n = e$. So, we extend this to real number $x$.

Exercise 1..32   Show $${\lim_{x \rightarrow -\infty}(1 + \frac{1}{x})^x = e}$$.

$x \to -\infty$ For $x \to -\infty$, it is better to put $y = -x$ to avoid negative infinity. Now $y \to \infty$ and by Example1.32

SOLUTION Put $x = -y$. Then

$$\lim_{x \to -\infty}(1 + \frac{1}{x})^x = \lim_{y \to \infty}(1 - \frac{1}{y})^{-y} = \lim_{y \to \infty}(\frac{y-1}{y})^{-y}$$

$$= \lim_{y \to \infty}(\frac{y}{y-1})^y = \lim_{y \to \infty}(1 + \frac{1}{y-1})^y$$

$$= \lim_{y \to \infty}(1 + \frac{1}{y-1})^{y-1}(1 + \frac{1}{y-1}) = e\ensuremath{ \blacksquare}$$

Exercise A

2.

From the table, find the following values:

(a) $${\log{20}}$$ (b) $${\log{16}}$$ (c) $${\log{3^{4}}}$$ (d) $${\log{0.1}}$$ (e) $${\log{\sqrt{630}}}$$ (f) $${\log{0.4}}$$

$$\begin{array}{ll\vert ll} n & \log{n} & n & \log{n} \ \hline... ... & 1.39 & 9 & 2.20\\ 5 & 1.61 & 10 & 2.30\ \hline \end{array}$$

3.

Solve the following equations:

(a) $${\log{x} = 2}$$ (b) $${\log{x} = -1}$$ (c) $${(2-\log{x})\log{x} = 0}$$

(d) $${\log{(2x+1)(x+2)} = 2\log(x+2)}$$

Exercise B

1.

Find the limit of the following functions:

(a) $${\lim_{x \rightarrow 0}\frac{\log{(1+x)}}{x}}$$ (b) $${\lim_{h \rightarrow 0}\frac{e^{h} - 1}{h}}$$ (c) $${\lim_{x \rightarrow a}\frac{\log{x} - \log{a}}{x - a}}$$

(d) $${\lim_{x \rightarrow 0}\frac{(1+x)^{\alpha} - 1}{x}}$$ (e) $${\lim_{x \rightarrow 0+}x^x}$$

5.

Given the arithmetic mean $a_{n+1} = \frac{a_{n} + b_{n}}{2}$ and the geometric mean $b_{n+1} = \sqrt{a_{n}b_{n}}$. Answer the following questions:

(a) Show $\{a_{n}\}$ and $\{b_{n}\}$ converge for $a_{0}, b_{0} \geq 0$.

(b) Show $\lim_{n \to \infty}a_{n} = \lim_{n \to \infty}b_{n}$. This limit is denoted by ${\rm agm}(a_{0},b_{0})$